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2001 Stanford Math Tournament Advanced Topics 1. Suppose sinh(x)= e x e -x 2 . What is the inverse function of sinh(x)? Solution: ln(x + x 2 + 1) y = e x e -x 2 . Switch x and y to find the inverse, so x = e y e -y 2 , or 2x = e y e y , so 2xe y = e 2y 1, and e 2y 2xe y 1 = 0. By using the quadratic formula, we see that e y = 2x± 4x 2 +4 2 = x ± x 2 + 1, so y = ln(x + x 2 + 1) (since y cannot be ln(x x 2 + 1), since x x 2 + 1 is always negative, and ln is not defined for negative numbers). 2. Write 20 10 7017! 1708! as a decimal in base 6. The subscript indicates the base in which the number is written (i.e., 20 10 is 20 base 10.) Solution: 0.1 6 or 0.1 The easiest way to solve this problem is to convert everything into base 10. 70 17 = 7 × 17 1 +0 × 17 0 = 119 170 8 = 1 × 8 2 +7 × 8 1 +0 × 8 0 = 120 20 10 70 17 ! 170 8 ! = 20 119! 120! = 1 6 Converting to base 6, 1 6 =0.1 6 . 3. There are 36 penguins in a row, and Barbara Manatee is standing in front of them. In general, a penguin rotation of penguins p 1 ,p 2 ,...,p n is a rearrangement of them such that p 1 moves to where p 2 was standing, and in general p i moves to where p i+1 was standing, and p n moves to where p 1 was standing. So, after a penguin rotation, the new order of these penguins is p n ,p 1 ,p 2 ,...,p n1 . Whenever Barbara Manatee blows her whistle, the 2-4 penguins go through a penguin rotation, the 5-9 penguins go through a penguin rotation, the 10-16 penguins go through a penguin rotation, the 17-25 penguins go through a penguin rotation, and the 26-36 penguins go through a penguin rotation. What is the least positive number of whistle blows such that the penguins all return to their original position? Solution: 3465 Since we have groups of 3,5,7,9,11 penguins rotating, for them all to be in the right place requires the number of whistle blows to be a multiple of the least common multiple of 3,5,7,9,11 (denoted [3, 5, 7, 9, 11]). So, we calculate [3, 5, 7, 9, 11] = [5, 7, 9, 11] = 5 · 7 · 9 · 11 = 3465. 4. Eleven pirates find a treasure chest. When they split up the coins in it, evenly among all the pirates, they find that there are 5 coins left. They throw one pirate overboard and split the coins again, only to find that there are 3 coins left over. So, they throw another pirate over and try again. This time, the coins split perfectly. What is the least number of coins there could have been? Solution: 423 Let n be the number of coins in the chest. Then, n 5 (mod 11) (this means that the remainder of n when divided by 11 is 5), n 3 (mod 10), and n 0 (mod 9). The Chinese Remainder Theorem gives an elegant solution to this problem, or you can solve it as follows. Since n 3 (mod 10), we know that the last digit of n must be 3. Also, since n 0 (mod 9), n is divisible by 9, so n =9q for some q. Since n ends in a 3, q must end in a 7. That means that q looks like 10p + 7, so n = 9(10p + 7) = 90p + 63. Now, you can just try p =1, 2,..., and you’ll find that p = 4 works. 5. Evaluate 1 · 2 1 +3 · 2 2 +5 · 2 3 +7 · 2 4 + ... Solution: 3

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Page 1: Smt Solutions

2001 Stanford Math Tournament

Advanced Topics

1. Suppose sinh(x) = ex−e−x2 . What is the inverse function of sinh(x)?

Solution: ln(x+√x2 + 1)

y = ex−e−x2 . Switch x and y to find the inverse, so x = ey−e−y

2 , or 2x = ey − e−y, so 2xey = e2y − 1,

and e2y − 2xey − 1 = 0. By using the quadratic formula, we see that ey = 2x±√4x2+42 = x±√x2 + 1,

so y = ln(x+√x2 + 1) (since y cannot be ln(x−√x2 + 1), since x−√x2 + 1 is always negative, and

ln is not defined for negative numbers).

2. Write 20107017!1708!

as a decimal in base 6. The subscript indicates the base in which the number is written(i.e., 2010 is 20 base 10.)

Solution: 0.16 or 0.1

The easiest way to solve this problem is to convert everything into base 10.

7017 = 7× 171 + 0× 170 = 119

1708 = 1× 82 + 7× 81 + 0× 80 = 120

20107017!

1708!= 20

119!

120!=

1

6

Converting to base 6, 16 = 0.16.

3. There are 36 penguins in a row, and Barbara Manatee is standing in front of them. In general, apenguin rotation of penguins p1, p2, . . . , pn is a rearrangement of them such that p1 moves to wherep2 was standing, and in general pi moves to where pi+1 was standing, and pn moves to where p1was standing. So, after a penguin rotation, the new order of these penguins is pn, p1, p2, . . . , pn−1.Whenever Barbara Manatee blows her whistle, the 2-4 penguins go through a penguin rotation, the5-9 penguins go through a penguin rotation, the 10-16 penguins go through a penguin rotation, the17-25 penguins go through a penguin rotation, and the 26-36 penguins go through a penguin rotation.What is the least positive number of whistle blows such that the penguins all return to their originalposition?

Solution: 3465

Since we have groups of 3,5,7,9,11 penguins rotating, for them all to be in the right place requiresthe number of whistle blows to be a multiple of the least common multiple of 3,5,7,9,11 (denoted[3, 5, 7, 9, 11]). So, we calculate [3, 5, 7, 9, 11] = [5, 7, 9, 11] = 5 · 7 · 9 · 11 = 3465.

4. Eleven pirates find a treasure chest. When they split up the coins in it, evenly among all the pirates,they find that there are 5 coins left. They throw one pirate overboard and split the coins again, onlyto find that there are 3 coins left over. So, they throw another pirate over and try again. This time,the coins split perfectly. What is the least number of coins there could have been?

Solution: 423

Let n be the number of coins in the chest. Then, n ≡ 5 (mod 11) (this means that the remainder ofn when divided by 11 is 5), n ≡ 3 (mod 10), and n ≡ 0 (mod 9). The Chinese Remainder Theoremgives an elegant solution to this problem, or you can solve it as follows. Since n ≡ 3 (mod 10),we know that the last digit of n must be 3. Also, since n ≡ 0 (mod 9), n is divisible by 9, son = 9q for some q. Since n ends in a 3, q must end in a 7. That means that q looks like 10p+ 7, son = 9(10p+ 7) = 90p+ 63. Now, you can just try p = 1, 2, . . ., and you’ll find that p = 4 works.

5. Evaluate 1 · 2−1 + 3 · 2−2 + 5 · 2−3 + 7 · 2−4 + . . .

Solution: 3

Page 2: Smt Solutions

We will express this as a sum of geometric series as follows:

1 · 2−1 + 3 · 2−2 + 5 · 2−3 + 7 · 2−4 + . . . =∞∑n=1

(2n− 1)2−n

=

∞∑n=1

(2n)2−n −∞∑n=1

2−n

=∞∑n=1

n2−n+1 − 1/2

1− 1/2

=

∞∑n=1

n∑j=1

2−n+1 − 1

=

∞∑j=1

∞∑n=j

2−n+1 − 1

=

∞∑j=1

2−j+1

1− 1/2− 1

=

∞∑j=1

2−j+2 − 1

=2

1− 1/2− 1

= 3

6. How many subsets of {n | n > 0 and n is a multiple of 3 less than 100} are also subsets of {n | n > 0and n is a multiple of 4 less than 100}?Solution: 256

Let A = {n | n > 0 and n is a multiple of 3 less than 100} and B = {n | n > 0 and n is a multiple of4 less than 100}. Then, A ∩ B consists of all multiples of 12 less than 100; there are 8 of these. Thesubsets of A that are also subsets of B are exactly the subsets of A ∩ B; since A ∩B has 8 elements,it has 28 = 256 subsets.

7. There are 2000 dots spaced evenly around a circle. If 4 distinct dots A, B, C, and D are pickedrandomly, what is the probability that AB intersects CD?

Solution: 13

Randomly pick the 4 dots A, B, C, and D. Now, if we start from A and go clockwise, there are 6equally likely possibilities for the order that the other 3 dots will appear:

B, C, D C, B, D D, B, CB, D, C C, D, B D, C, B

AB and CD intersect if and only if B is the second of these three dots; this happens with probability13 .

8. Ashley, Bob, Carol, and Doug are rescued from a desert island by a pirate who forces them to play agame. Each of the four, in alphabetical order by their first names, is forced to roll two dice. If the totalon the two dice is either 8 or 9, the person rolling the dice is forced to walk the plank. The players goin order until one player loses: A, B, C, D, A, B, .... What is the probability that Doug survives?

Solution: 148/175

Page 3: Smt Solutions

We will instead find the probability that Doug doesn’t survive. Notice that the only time Doug doesn’tsurvive is if all four survive the first n rounds (where a round is once through all four), and then Alice,Bob, and Carol survive the n+ 1-st round but Doug does not. The probability of rolling an 8 or 9 is536 +

436 = 1

4 , so the probability that all four survive a round is(34

)4= 81256 . So, all four survive the

first n rounds with probability(81256

)n. The probability that Alice, Bob, and Carol survive the n+1-st

round but Doug does not is(34

)3 14 =

27256 . So, the probability that Doug has to walk the plank in the

n+ 1-st round is 27256

(81256

)n. Thus, the probability that Doug doesn’t survive is

∞∑n=0

27

256

(81

256

)n=

27

256

1

1− 81256

=27

175.

So, the probability that Doug survives is 148175 .

9. There are 19 men numbered 1 though 19 and 20 women numbered 1 through 20 entered in a computerdating service. The computer wants to match every man to a compatible woman, and each man isonly compatible with women who have a number that is greater than equal to his (i.e. man 19 is onlycompatible with women 19 and 20, man 18 is only compatible with women 18, 19, 20, etc.). If eachwomen is matched with at most one man, let n be the number of ways that the computer can matchthem. What is the prime factorization of n?

Solution: 219

There are 2 ways to match man 19, with either woman 19 or woman 20. Given that, there are 2 waysto match man 18, with either woman 18 or with the one that man 19 didn’t choose. This continues allthe way back to man 1, so there are 2 choices for each of the 19 men, so the total number of matchingsis 219.

10. David is playing with Legos with velcro attached to the ends. He has green Legos of length 1, blueLegos of length 2, and red Legos of length 3, and wants to combine them (by attaching them at theends) to make a “super-lego” of length 10. If any different ordering of colors is considered a distinct“super-lego”, how many ways can he make this “super-lego”?

Solution: 274

Let an be the number of ways to make a super-lego of length n. Then a1 is of course 1, a2 = 2 (wecan have either a lego of length 2 or 2 legos of length 1), and a3 = 4. Now, observe that to get asuper-lego of length n, we must have either taken a super-lego of length n− 3 and attached a lego oflength 3, a super-lego of length n− 2 and attached a lego of length 2, or a super-lego of length n− 1and attached a lego of length 1. So, an = an−3+ an−2+ an−1. Using this relationship, it’s easy to findthat a10 = 274.

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2001 Stanford Math Tournament

Algebra

1. Find the result of adding seven to the result of forty divided by one-half.

Solution: 87

Since we are dividing forty by one-half, the result is 80 (not 20), and adding 7 gives 87.

2. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all threevalves open, the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours, and with onlyvalves B and C open it takes 2 hours. How many hours will it take to fill the tank with only valves Aand B open?

Solution: 1.2 = 65

Let rA, rB , rC be the rates of valvesA, B, and C in tanks per hour, respectively. Then, 1(rA+rB+rC) =1, 1.5(rA + rC) = 1, and 2(rB + rC) = 1. So, rA + rB + rC = 1, rA + rC = 2

3 , and rB + rC = 12 . So,

rA + rB = 2(rA + rB + rC)− (rA + rC)− (rB + rC) = 2− 23 − 12 = 56 . Thus, it takes

65 = 1.2 hours to

fill the tank with only valves A and B open.

3. Julie has a 12 foot by 20 foot garden. She wants to put fencing around it to keep out the neighbor’sdog. Normal fenceposts cost $2 each while strong ones cost $3 each. If Julie needs one fencepost forevery 2 feet and has $70 to spend on fenceposts, what is the greatest number of strong fenceposts shecan buy?

Solution: 6

Let x be the number of normal fenceposts and y be the number of strong fenceposts. The perimeterof Julie’s garden is 64 feet, and she needs one fencepost every 2 feet, so x + y = 32. The total cost ofthe fenceposts is 2x+ 3y, and we want 2x+ 3y ≤ 70. Since 2x+ 2y = 64, we find that y ≤ 6.

4. p(x) is a real polynomial of degree at most 3. Suppose there are four distinct solutions to the equationp(x) = 7. What is p(0)?

Solution: 7

Since the degree of p(x)− 7 is at most 3, it has at most 3 roots. But we have 4 solutions, so p(x) − 7must be constant and equal to zero. So, p(x) = 7, and thus p(0) = 7.

5. Let f : N→ N be defined by f(x) =

{2, x = 0(f(x− 1))2, x = 0

What is log2 f(11)?

Solution: 2048

For x = 0, log2 f(x) = 2 log2 f(x − 1). For x = 0, log2 f(x) = 1. So, in general, log2 f(x) = 2x. Thus,log2 f(11) = 211 = 2048.

6. If for three distinct positive numbers x, y, and z,

y

x− z=

x+ y

z=

x

y,

then find the numerical value of x/y.

Solution: 2

Since yx−z = x+y

z, yz = x2 + xy − xz − yz. Since y

x−z = xy, y2 = x2 − xz. Finally, since x+y

z= xy,

xz = yx+ y2 = yx+ x2 − xz = yz + yz = 2yz. So, x/y = 2.

7. If logAB + logB A = 3 and A < B, find logB A.

Solution: 3−√5

2

Let x = logB A. Then Bx = A, so B = A1x . Thus, logAB = 1

x . So, x + 1x = 3. Multiplying through

by x, x2 + 1 = 3x, so x2 − 3x+ 1 = 0. By the quadratic formula, x = 3±√52 Since A < B, logB A < 1,

so x = 3−√52

Page 5: Smt Solutions

8. Determine the value of 1 + 12+ 1

1+ 12+ 11+···

.

Solution: 1+√3

2

Let x = 1 + 12+ 1

1+ 1

2+ 11+···

. Then, observe that x = 1 + 12+ 1x

, so x = 1 + x2x+1 = 3x+1

2x+1 . Therefore,

2x2+ x = 3x+1, so 2x2− 2x− 1 = 0. By the quadratic formula, x = 2±√4+84 = 2±2√3

4 = 1±√32 . Now,

1 + 12+ 1

1+ 1

2+ 11+···

is clearly positive, so it must be 1+√3

2 .

9. Find all solutions to (x− 3)(x− 1)(x+ 3)(x+ 5) = 13.

Solution: ±√3− 1,±√17− 1Let y = x+1, so x = y−1, and our new equation is (y−4)(y−2)(y+2)(y+4) = 13, or (y2−16)(y2−4) =13, and by expanding, we get that y4 − 20y2 + 51 = 0, so (y2 − 3)(y2 − 17) = 0, so y = ±√17,±√3,and x = ±√3− 1,±√17− 1.

10. Suppose x, y, z satisfy

x+ y + z = 3

x2 + y2 + z2 = 5

x3 + y3 + z3 = 7

Find x4 + y4 + z4.

Solution: 9

We have

9 = (x+ y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz) = 5 + 2(xy + yz + xz)

so xy + yz + xz = 2. Similarly,

15 = (x2 + y2 + z2)(x + y + z)

= x3 + y3 + z3 + x2y + y2x+ y2z + z2y + z2x+ x2z

= 7 + xy(x+ y) + yz(y + z) + zx(z + x)

= 7 + xy(3− z) + yz(3− x) + zx(3− y) [since x+ y + z = 3]

= 7 + 3(xy + yz + zx)− 3xyz

= 13− 3xyz, since xy + yz + zx = 2

So, xyz = −23 . Finally,

21 = (x3 + y3 + z3)(x+ y + z)

= x4 + y4 + z4 + x3y + y3x+ y3z + z3y + z3x+ x3z

= x4 + y4 + z4 + xy(x2 + y2) + yz(y2 + z2) + zx(z2 + x2)

= x4 + y4 + z4 + xy(5− z2) + yz(5− x2) + zx(5− y2)

= x4 + y4 + z4 + 5(xy + yz + zx)− xyz(x+ y + z)

= x4 + y4 + z4 + 12

So, x4 + y4 + z4 = 9.

Page 6: Smt Solutions

2001 Stanford Math Tournament

Calculus

1. What is d2959

dx2959sinx?

Solution: − cosxNotice that the fourth derivative of sinx is again sinx, and 2959 = 4 × 739 + 3. So, the 2959-thderivative of sinx is the same as the third derivative of sinx, which is just − cosx.

2. If f(x) = [x] is the greatest integer function, what is f ′′′(3.7)?

Solution: 0

Notice that f is constant on the interval (3, 4), so f ′ = 0 on (3, 4). Thus, f ′′′ = 0 on (3, 4), sof ′′′(3.7) = 0.

3. Suppose that f is a monotonically increasing continuous function defined on the real numbers. We

know that f(0) = 0 and f(2) = 3. Let S be the set of all possible values of∫ 20 f(x) dx. What is the

least upper bound of S?

Solution: 6

Since f is monotonically increasing, we know that for all x ≤ 2, f(x) ≤ 3. Also, for all x ≥ 0, f(x) ≥ 0.

So, 6 is an upper bound of∫ 20f(x) dx. Now, we will show that it is the least upper bound. Consider

functions fn defined by

fn(x) =

{3nx, x ≤ 1

n

3, x > 1n

Then,∫ 20 f(x) dx =

∫ 1n

0 3nx dx +∫ 21n3 dx = 3

2n + 6 − 3n = 6 − 3

2n , so we have found monotonically

increasing functions whose integrals are arbitrarily close to 6. Thus, the least upper bound of∫ 20f(x) dx

is 6.

4. Evaluate∫ 5−4

x2

|x|dx.

Solution: 412= 20.5 = 201

2

For x ≤ 0, |x| = −x. For x ≥ 0, |x| = x. So,

∫ 5−4

x2

|x| dx =

∫ 0−4

x2

−x dx+∫ 50

x2

xdx =

∫ 0−4−xdx+

∫ 50

xdx

= −12x2∣∣∣∣0

−4+

1

2x2∣∣∣∣5

0

= −0 + 1

2· 16 + 1

2· 25− 0

= 8 +25

2

and this is 412 .

5.∫∞1

1x2+3x =?

Solution: 13ln 4 = ln 4

3

Solution: The integral given is equal to 13 limu→∞∫ u1

(1x− 1x+3

)= 13 limu→∞ (lnu− (ln(u + 3)− ln 4))

= 13 ln 4.

6. Given a point (p, q) on the ellipse x2

a2+ y2

b2= 1 where p = 0, find the x-intercept of the tangent line at

(p, q) in terms of p, a, and b. (Note that a, b = 0.)

Solution: a2

p

Page 7: Smt Solutions

If q = 0, then we are either at the point (a, 0) or (−a, 0), so in either case, we see that a2

p gives thecorrect answer. Otherwise, suppose q > 0. Then, we are in the top half of the ellipse, which has

equation y =√b2 − x2b2

a2. Remember that the slope of the tangent line at (p, q) is just dy

dxat (p, q).

By the chain rule, dydx

= − 12(b2 − x2b2

a2

)−1/22xb2

a2= − xb2

ya2. So, at (p, q), dy

dx= − pb2

qa2. This means that

the tangent line at (p, q) looks like y = − pb2qa2

x + c for some c. We know that (p, q) is a point on the

line, so q = − pb2qa2

p + c, and this tells us that c = q + pb2

qa2p. So, the equation of the tangent line is

y = − pb2qa2

x+ q + pb2

qa2p.

So, the x-intercept of the tangent line is the x satisfying 0 = − pb2qa2 x + q + pb2

qa2 p. Solving this linear

equation, we find that x = q2a2+p2b2

pb2. However, this is not yet in terms of p, a, and b. Remember that

(p, q) is a point on the elliptic curve, so p2

a2+ q2

b2= 1. This means that p2b2+ q2a2 = a2b2; substituting

this into the numerator of x, we get x = a2b2

pb2= a2

p.

Now, notice that the tangent line to (p,−q) is just the tangent line of (p, q) reflected over the x-axis,so it has the same x-intercept.

7. Find the number of real solutions to sin(6πx) = x, where x is in radians.

Solution: 11

There is one solution at x = 0. Now, consider the graph of sin(6πx)−x. At 112 , sin(6πx) = sin(π/2) = 1,so sin(6πx) − x > 0. At 3

12 , sin(6πx) − x < 0, so there is a solution to sin(6πx) = x between 112 and

312 . Similarly, there are solutions between 3

12 and 512 ,

512 and 7

12 ,712 and 9

12 , and912 and 11

12 , giving5 solutions with x positive. There is not a solution between 11

12 and 1, since sin(6π · 1112 ) = −1 andsin(6π) = 0, and there are no solutions for x > 1, since then x > 1 > sin(6πx). Since the graph ofsin(6πx) − x is symmetric around the origin, there must be 5 negative solutions as well, for a total of11 solutions.

8. If a, b, and c are positive real numbers such that a+ b + c = 16 and a2 + b2 + c2 = 160, what is themaximum possible value of abc?

Solution: 12827(7√7− 10)

First, notice that a+b = 16−c and a2+b2 = 160−c2, so 2ab = (a+b)2−(a2+b2) = (16−c)2−(160−c2) =2c2 − 32c+ 96, and ab = c2 − 16c+ 48. Then, abc = c3 − 16c2 + 48c. To find the maximum, we wantthe derivative of c3 − 16c2 + 48c to be 0, so we want 3c2 − 32c+ 48 = 0. Using the quadratic formula,

we find that c = 16±4√73 . If c = 16+4

√7

3 , then abc < 0, which is clearly not the maximum. So, we take

c = 16−4√73 . Substituting into abc = c3 − 16c2 + 48c, we find that abc = 128

27 (7√7− 10).

9. For a given k > 0, n ≥ 2k > 0, consider the square R in the plane consisting of all points (x, y) with0 ≤ x, y ≤ n. Color each point in R gray if xy

k≤ x+ y, and blue otherwise. Find the area of the gray

region in terms of n and k.

Solution: 2k(n+ k ln n−k

k

)First, observe that xy

k≤ x+y iff xy ≤ kx+ky iff xy−kx−ky ≤ 0 iff (x−k)(y−k) = xy−kx−ky+k2 ≤ k2.

So, we really want to find the area of the part of (x−k)(y−k) ≤ k2 that is contained in R. Notice thatthe graph (x − k)(y − k) = k2 is just a hyperbola, and the bottom half of the hyperbola lies entirelyoutside the first quadrant, so we only need to look at the top half of the hyperbola.

(k, 0)

Page 8: Smt Solutions

The graph of the top half is y = k2

x−k + k, which is strictly decreasing. Notice that k2

x−k + k = n for

x = k2

n−k + k = nkn−k , so

k2

x−k + k ≥ n for all x ≤ nkn−k . Thus, we can express the gray area as

∫ nkn−k

0

n dx+

∫ nnkn−k

(k2

x− k+ k

)dx =

n2k

n− k+ (k2 ln(x − k) + kx)|nnk

n−k

=n2k

n− k+ [k2 ln(n− k) + kn]− (k2 ln

k2

n− k+ k

nk

n− k)

= 2kn+ k2 ln(n− k)− k2 lnk2

n− k

= 2kn+ k2 ln(n− k)− 2k2 ln k + k2 ln(n− k)

= 2k

(n+ k ln

n− k

k

)

10. Let f(x) =(x2 − 1

)n, where n is a positive integer. Determine, in terms of n, (a, b, c), where a,b,

and c are the number of distinct roots of f (n)(x) in the intervals (−∞,−1), (−1, 1), and (1,+∞),respectively.

Solution: (0, n, 0)

A polynomial of degree k has at most k roots and differentiating a polynomial decreases the multiplicityof each existing root by 1. Also, the derivative of any differentiable function with two different real 0shas a real 0 strictly between them by Rolle’s theorem (the mean value theorem). This means that for0 ≤ k ≤ n, the kth derivative of f has roots of multiplicity n−k at -1 and 1 and k roots of multiplicity1 in (−1, 1). So, for k = n, we see that there are roots of multiplicity 0 (i.e. no root) at -1 and 1, andn roots in (−1, 1). Since this polynomial is of degree n, these must be the only roots.

Page 9: Smt Solutions

2001 Stanford Math Tournament

Geometry

1. Find the coordinates of the points of intersection of the graphs of the equations y = |2x| − 2 andy = −|2x|+ 2.

Solution: (1, 0), (−1, 0)(x, y) is a point of intersection iff y = |2x|−2 = −(|2x|−2), so y must be 0. Then, |2x| = 2, so x = ±1.

2. Jacques is building an igloo for his dog. The igloo’s inside and outside are both perfectly hemispherical.The interior height at the center is 2 feet. The igloo has no door yet and contains 254

2187π cubic yardsof hand-packed snow. What is the circumference of the igloo at its base in feet?

Solution: 143π

Let R be the radius of the igloo at its base, in feet. Then, the amount of snow in the igloo is, in cubicfeet, 12

(43πR

3 − 43π · 23)= 23π(R

3 − 8). This is equal to 2542187π cubic yards, or 25481 = 33 · 2542187 cubic

feet. Therefore, 23π(R3 − 8) = 254

81 π, so R3 = 34327 , and R = 7

3 . So, the circumference of the igloo is2πR = 14

3 π feet.

3. Find the area of the convex quadrilateral whose vertices are (0, 0), (4, 5), (9, 21), (−3, 7).Solution: 165

2

We can split this into two triangles, one with vertices (0, 0), (4, 5), (9, 21) and the other with vertices(0, 0), (9, 21), (−3, 7). To find the area, we will first find a formula for the area of a triangle with vertices(0, 0), (a, b), (c, d) and then apply this formula to our two triangles.

(c, d)

(a, b)

A

C

B

As the figure demonstrates, we can find the area of the triangle by first finding the area of the largetriangle (the one with vertices (0, 0), (a, b), (a, 0)) and then subtracting the area of the regions A, B,and C. The large triangle clearly has area cd2 . Region A has area ab2 , region B has area (c − a)b, and

region C has area (c−a)(d−b)2 . So, the area of the triangle is cd2 − ab2 − (c − a)b − (c−a)(d−b)

2 = ad−bc2 .

Of course, the area should only be positive, so we need to take the absolute value of this. (If you’refamiliar with determinants, think about what this means in terms of determinants.)

Using the formula, the area of the triangle with vertices (0, 0), (4, 5), (9, 21) is |4·21−5·9|2 = 392 and the

area of the triangle with vertices (0, 0), (9, 21), (−3, 7) is |9·7+21·3|2 = 63. So, the total area of thequadrilateral is 1652 .

4. E is a point in the interior of rectangle ABCD. AB = 6, triangle ABE has area 6, and triangle CDEhas area 12. Find (EA)2 − (EB)2 + (EC)2 − (ED)2.

Solution: 0

Let l be the line through E perpendicular to AB and CD. Let F and G be the intersections of l withAB and CD, respectively. Then, AF = DG, call this a, and BF = CG, call this b. Since triangleABE has area 6, EF = 2. Since triangle CDE has area 12, EG = 4.

A B

CD

E

F

G

a b

a b

2

4

Page 10: Smt Solutions

By the Pythagorean Theorem, we have the following equalities:

a2 + 4 = (EA)2, b2 + 4 = (EB)2, b2 + 16 = (EC)2, a2 + 16 = (ED)2

Thus, (EA)2 − (EB)2 + (EC)2 − (ED)2 = 0.

5. Two identical cones, each 2 inches in height, are held one directly above another with the pointed endfacing down. The upper cone is completely filled with water. A small hole is punctured in the bottomof the upper cone so that the water trickles down into the bottom cone. When the water reaches adepth of 1 inch in the bottom cone, what is its depth in the upper cone?

Solution: 3√7

Let r be the radius of the base of one cone. Then, the amount of water is V = 132πr

2 = 23πr

2. Whenthe height of the water in the bottom cone is 1, using similar triangles, the volume of water in the

bottom cone is Vb =131π(r2

)2= 112πr

2. So, the volume of water is the upper cone is Vu = V − Vb =712πr

2 = 13hnewπr

2new where hnew is the depth of the water in the upper cone and rnew is the radius of

the cone formed by the remaining water. Again, by similar triangles, rnewr

= hnew2 . Substituting,

Vu =7

12πr2 =

1

3πhnew

(rhnew

2

)2.

Solving this equation, we find that hnew = 3√7.

6. Find the radius of a circle inscribed in the triangle determined by the lines 4x+ 3y = 24, 56x− 33y =−264, and 3x− 4y = 18.

Solution: 4

First, observe that the lines 4x + 3y = 24 and 3x − 4y = 18 are perpendicular, since their slopesare negative reciprocals. We can also find their intersection point; 4x + 3y = 24 can be rewritten asy = 24−4x

3 , while 3x − 4y = 18 can be rewritten as y = 3x−184 . Setting these two equal, we have

24−4x3 = 3x−18

4 . Cross-multiplying, 4(24− 4x) = 3(3x − 18), so 96 − 16x = 9x − 54, and 150 = 25x,and x = 6. Then, y = 24−4x

3 = 0. Thus, one vertex of the triangle is (6, 0). Similarly, we find that theother two vertices of the triangle are (0, 8) and (− 665 ,− 725 ).Now, we must find the lengths of the legs of the triangle. The leg between the vertices (6, 0) and (0, 8)has length

√(6− 0)2 + (0− 8)2 = 10, while the other leg has length 24. Thus, the hypotenuse of the

triangle has length√102 + 242 = 26.

Let the radius of the incircle be x. Using the fact that the tangents from a point to a circle have thesame length, we have the following diagram:

10−x

24−x

24−x

10−x

xx

x

Thus, (10− x) + (24− x) = 26, so x = 4.

7. In the figure, AB is tangent at A to the circle with center O; point D is interior to the circle; and DBintersects the circle at C. If BC = DC = 3, OD = 2 and AB = 6, then find the radius of the circle.

Page 11: Smt Solutions

O

A B

D

C

Solution:√22

Let θ = ∠ODB and r be the radius of circle O.

O

A B6

3

32D

C

θ

r

Observe that ABO is a right triangle, so by the Pythagorean theorem. BO2 = 36+ r2. By the law ofcosines with triangle ODC, r2 = 4 + 9 − 12 cos θ, so −12 cosθ = r2 − 13. By the law of cosines withtriangle ODB, BO2 = 4 + 36− 24 cosθ = 40 + 2(r2 − 13) = 14 + 2r2. Therefore, 36 + r2 = 14 + 2r2,so r2 = 22, and r =

√22.

8. Let S be the solid tetrahedron with boundary points (0, 0, 0), (2, 4, 0), (5, 1, 0), (3, 2, 10). Let z1 =max{q | (125 , 2310 , q) ∈ S} and let z2 = max{r | (92 , 54 , r) ∈ S}. Find z1 − z2.

Solution: 154

A tetrahedron is made up of four planes. Notice that z1 and z2 must lie on the boundary of S, so eachlies on one of the four planes. However, we can ignore the plane defined by (0, 0, 0), (2, 4, 0), (5, 1, 0)because z1 and z2 will not be on this plane (as it is lower than all the rest, and we want the maximum).Also notice that a vertical line intersects each of the other 3 planes exactly once, but only the lowestintersection point is actually on the tetrahedron.

To find the planes, recall that any plane that doesn’t go through the origin can be represented asax + by + cz = 1. Let P1 be the plane defined by (2, 4, 0), (5, 1, 0), (3, 2, 10). Then, 2a + 4b = 1,5a + b = 1, and 3a + 2b + 10c = 1. Solving, we find that P1 is the plane 1

6x + 16y + 1

60z = 1.A plane that does go through the origin can be written as x + by + cz = 0. Let P2 be the planedefined by (0, 0, 0), (2, 4, 0), (3, 2, 10). Then, using the same method as above, we find that the planeis x − 1

2y − 15z = 0. Let P3 be the plane defined by (0, 0, 0), (5, 1, 0), (3, 2, 10). Then, it has equation

x− 5y + 710z = 0.

Using these equations, we find that (125 ,2310 , 13) is on P1, (

125 ,2310 ,

254 ) is on P2, and (125 ,

2310 , 13) is on P3.

Only the lowest is on the tetrahedron, so z1 =254 . Similarly, we find that z2 =

52 , so z1 − z2 =

154 .

9. Circles A and B are tangent and have radii 1 and 2, respectively. A tangent to circle A from thepoint B intersects circle A at C. D is chosen on circle B so that AC is parallel to BD and the twosegments BC and AD do not intersect. Segment AD intersects circle A at E. The line through B andE intersects circle A through another point F . Find EF .

Page 12: Smt Solutions

Solution: 2√3/3

Draw a line from A perpendicular to BD. Let G the point of intersection of this line with BD. Now,notice that ACBG is a rectangle.

A B

D

E

F

C

G

Since C is on circle A, AC = 1. Also, AB = 3 since circle A has radius 1 and circle B has radius 2.By the pythagorean theorem, BC = AG = 2

√2. Then, AGD is a right triangle with legs 1 and 2

√2,

so AD = 3. Then, since A, E, and D are collinear, AD = AE + ED, so ED = 2. This means thatED = BD = 2, so DEB is an isosceles triangle with vertex D. Notice that AEF is also isosceles, sinceE and F are both on circle A. Also, ∠AEF = ∠BED, so the triangles AEF and DEB are similar.In particular, the ratio of a side of DEB to a side of AEF is 2 to 1 because BD = 2 and AF = 1, soFE = 1

2EB, and FE = 13BF .

So, now we just need to find BF . Let θ = ∠AFE. Then, ∠AEF = θ, so ∠FAE = 180◦−2θ. Also, sinceAEF and BED are similar, ∠BDE = 180◦ − 2θ. Since AGD is a right triangle, ∠GAD = 2θ − 90◦,so ∠GAF = ∠GAD + ∠FAE = (2θ − 90◦) + (180◦ − 2θ) = 90◦. Since CAG is also a right angle, thismeans that C, A, and F are collinear. So, BCF is a right triangle, and it has legs 2 and 2

√2. So,

BF = 2√3, and FE = 1

3BF = 2√33 .

10. E is a point inside square ABCD such that ∠ECD = ∠EDC = 15◦. Find ∠AEB.

Solution: 60◦

We will use symmetry to solve this problem. Place points F, G, and H inside the square so thattriangles EDC, FCB, GBA, and HAD are congruent.

A B

CD

E

FH

G

Then trianglesEDH , FCE, GBF , andGAH are isosceles with vertex angle 60◦, so they are equilateral.So, GHEF is a rhombus. Then, since ∠DEC is 150◦ and ∠DEH and ∠CEF are both 60◦, ∠HEFis 90◦. By symmetry, the other angles of GHEF are also 90◦, so GHEF is a square.

Since AHG is equilateral and GHEF is a square, AEH is an isosceles triangle with vertex angle 150◦.Thus, ∠AEH is 15◦. Similarly, ∠BEF is 15◦, so ∠AEB is 60◦.

Page 13: Smt Solutions

2001 Stanford Math Tournament

General Test

1. If a & b is defined as 2a− ba, what value is associated with 3 & 2?

Solution: −23 & 2 = 2 · 3− 23 = 6− 8 = −2.

2. Find the coordinates of the points of intersection of the graphs of the equations y = |2x| − 2 andy = −|2x|+ 2.

Solution: (1, 0), (−1, 0)(x, y) is a point of intersection iff y = |2x|−2 = −(|2x|−2), so y must be 0. Then, |2x| = 2, so x = ±1.

3. ABCD is a 4-digit number. What is the largest number that can be formed with AB a prime 2-digitnumber and C and D each a prime 1-digit number?

Solution: 9777

The largest 2-digit prime number is 97, and the largest 1-digit prime is 7, so the largest number withthe specified properties is 9777.

4. Let A be the set of all non-composite positive integers. Let B be the set of all squares of integers. LetC be the set of all multiples of 3. What is the size of A ∩ (B ∪ C)?Solution: 2

B ∪ C = all squares and multiples of 3, so A ∩ (B ∪ C) = {1, 3}, and the size of this is 2.

5. What is i2959?

Solution: −iNotice that i2 = −1, i3 = −i, and i4 = 1. Since 2959 = 4× 739 + 3, i2959 = (i4)739i3 = i3 = −i.

6. The points Q = (9, 14) and R = (a, b) are symmetric with respect to the point P = (5, 3). What arethe coordinates of point R?

Solution: (1,−8)P is the midpoint of the segment QR, so a+9

2 = 5 and b+142 = 3. Thus, a = 1 and b = −8, so

R = (1,−8).7. If F (x) = 3x3 − 2x2 + x− 3, find F (1 + i).

Solution: −8 + 3iNotice that (1 + i)2 = 2i, so (1 + i)3 = 2i − 2. So, 3(1 + i)3 − 2(1 + i)2 + (1 + i) − 3 = 3(2i − 2) −2(2i) + (1 + i)− 3 = −8 + 3i.

8. Express the absolute value of the difference between 0.36 and 0.36 as a common fraction.

Solution: 1/275

Let x = 0.36. Then, 100x = 36.36, so 99x = 36, and x = 3699 =

411 . So, the difference between 0.36 and

0.36 is 411 − 9

25 =1275 .

9. Jonathan chooses 10 cards without replacement from a standard 52-card deck of cards (without jokers).What is the probability that he does not draw the 3 of clubs and he does not draw the King of spades?

Solution: 287442

The number of ways to draw 10 cards that are not either one of those cards is(5010

), and the total

number of ways to draw 10 cards is(5210

). So, the probability is

(5010)(5210)

=50!10!40!52!10!42!

= 42·4152·51 =

287442 .

Page 14: Smt Solutions

10. Julie has a 12 foot by 20 foot garden. She wants to put fencing around it to keep out the neighbor’sdog. Normal fenceposts cost $2 each while strong ones cost $3 each. If Julie needs one fencepost forevery 2 feet and has $70 to spend on fenceposts, what is the largest number of strong fenceposts shecan buy?

Solution: 6

Let x be the number of normal fenceposts and y be the number of strong fenceposts. The perimeterof Julie’s garden is 64 feet, and she needs one fencepost every 2 feet, so x + y = 32. The total cost ofthe fenceposts is 2x+ 3y, and we want 2x+ 3y ≤ 70. Since 2x+ 2y = 64, we find that y ≤ 6.

11. Anne has a cube that she wants to paint. If she decides to paint each face a different color and she has6 colors, how many distinct ways can she paint the cube? (If one painted cube can be rotated to getanother, they are the same.)

Solution: 30

Let’s label the colors 1 through 6. Anne wants to paint each face a different color, so she will use eachcolor exactly once. Since rotated paintings are the same, let’s always rotate our painting so that color1 is on the bottom. Then, there are 5 colors, and 5 faces left. She can color these in 5! = 120 ways.However, notice that she can spin the cube around (always keeping color 1 on the bottom), and thesepainting are all the same. There are 4 such rotations, so there are actually only 1204 = 30 ways to paintthe cube.

12. Find, in degrees, the sum of angles 1, 2, 3, 4, and 5 in the star-shaped figure shown.

1

3

45

2

Solution: 180◦

Let’s label the points where the star meets the circle A, B, C, D, and E.

1

3

45

2

A

B

C

DE

The intercepted arc of angle 1 is CD, so the measure of angle 1 is half the arc-length of CD. We cando the same for the other angles, so the sum of the angles is half the sum of the arc lengths AB, BC,CD, DE, and EA. These make up the whole circle, so their sum is 360◦. Therefore, the sum of angles1, 2, 3, 4, and 5 is 180◦.

13. Each valve A, B, and C, when open, releases water into a tank at its own constant rate. With all threevalves open, the tank fills in 1 hour, with only valves A and C open it takes 1.5 hours, and with onlyvalves B and C open it takes 2 hours. How many hours will it take to fill the tank with only valves Aand B open?

Solution: 1.2 = 65

Let rA, rB , rC be the rates of valvesA, B, and C in tanks per hour, respectively. Then, 1(rA+rB+rC) =1, 1.5(rA + rC) = 1, and 2(rB + rC) = 1. So, rA + rB + rC = 1, rA + rC = 2

3 , and rB + rC = 12 . So,

rA + rB = 2(rA + rB + rC)− (rA + rC)− (rB + rC) = 2− 23 − 12 = 56 . Thus, it takes

65 = 1.2 hours to

fill the tank with only valves A and B open.

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14. Lattice paths are paths consisting of one unit steps in the positive horizontal or vertical directions.How many distinct lattice paths are there from the origin to the point (5, 4)?

Solution: 126

To get to (5, 4) from the origin, you must go to the right 5 times and up 4 times. The four verticalsteps can be taken in any order in the 9 total steps, and thus there are

(94

)= 126 distinct lattice paths.

15. Three circles, each of area 4π, are all externally tangent. Their centers form a triangle. What is thearea of the triangle?

Solution: 4√3

Each circle has radius 2, so the triangle is equilateral with side length 4. Thus, the area of the triangleis 4√3.

16. Two equilateral triangles sharing an edge have a combined area of π. What is the square of the lengthof their shared edge?

Solution: 2π√3

3

Let s be the length the shared edge. An equilateral triangle of side length s has area s2√3

4 , so 2 s2√3

4 = π,

and s2 = 2π√3

3 .

17. Frogger wants to cross a stream. He starts on one bank and jumps onto the first passing log, which istraveling at 9 feet per second. The following logs are at speeds such that the second is twice is fast asthe first, the third is twice as fast as the second, the fourth is one-third as fast as the third, and thelast (fifth) is one-third as fast as the fourth (all of the logs travel in the same direction). If there is a2 second interval between jumps, then how far does he travel down river from his original point, oncehe reaches the opposite bank?

Solution: 158

The first log travels at 9 feet per second, so the second travels at 18 feet per second, the third travelsat 36 feet per second, the fourth travels at 12 feet per second, and the last travels at 4 feet per second.Frogger spends 2 seconds on each log, so he travels 2(9 + 18 + 36 + 12 + 4) = 158 feet.

18. Rice tuition was $7200 one semester. Each student takes a minimum of 12 credits and a maximum of20 credits. For each credit, a class meets for 3 hours a week. A semester is 15 weeks with no breaks.What is the difference in cost in dollars per hour of class that a student taking the maximum load anda student taking the minimum load pays, to the nearest penny?

Solution: $5.33

A student taking the minimum load has 12 · 3 · 15 = 540 class hours over the semester so pays7200540 = 40

3 = $13.33 per class hour. A student taking the maximum load has 20 · 3 · 15 = 900 classhours so pays 7200900 = $8 per class hour. So, the difference is $13.33− $8 = $5.33.

19. Suppose a1, a2, a3, . . . is a sequence of numbers such that a1 = 1 and an+1 = an + (2n + 1) for allpositive integers n. Find a20.

Solution: 400

In general, an = n2. This is certainly true for a1, and if an = n2, then an+1 = an + (2n + 1) =n2 + 2n+ 1 = (n+ 1)2, so it is true for all n (this sort of proof is called induction).

20. My spouse and I have 9 kids. Each child gets married and has exactly 9 children of their own. By thetime I have 100,000 descendants (including every generation) what is the longest title that applies tome? (Assuming no married blood relatives.)

a. grandparent

b. great grandparent

c. great great grandparent

Page 16: Smt Solutions

d. great great great grandparent

e. great great great great grandparent

f. great great great great great grandparent

g. great great great great great great grandparent

Solution: great great great great grandparent

I have 9 children, 92 grandchildren, 93 great grandchildren, etc. So, the number of descendants I have

in the n-th generation is 9 + 92 + . . .+ 9n = 9n+1−18 − 1. The smallest value of n for which this is at

least 100,000 is 6, which makes me a great great great great grandparent.

21. Find the greatest integer x for which 320 > 32x.

Solution: 6

First, notice that 32x = 25x, so 320 > 32x if and only if 34 > 2x. 34 = 81, and it is easy to check that26 is the largest power of 2 less than 81.

22. If x+ y = xy, with x and y real, what value can x not have?

Solution: 1

x+ y = xy if and only if xy − x− y + 1 = 1, or (x− 1)(y − 1) = 1. This is impossible if x = 1.

23. Instead of using two standard cubical dice in a board game, three standard cubical dice are used sothat the game goes more quickly. In the regular game, doubles are needed to get out of the “pit”. Inthe revised game, doubles or triples will get you out. How many times as likely is it for a player to getout of the “pit” on one toss under the new rules as compared to the old rules?

Solution: 8/3

The probability of rolling two different numbers with two different dice is 6·56·6 =56 , so the probability

of rolling doubles is 1− 56 = 16 . The probability of rolling three different numbers with three different

dice is 6·5·46·6·6 =59 , so the probability of rolling doubles or triples with three dice is 1− 5

9 =49 . So, the

ratio of the two is 49/16 =

83 .

24. A circle is drawn with center at the origin and radius 2.5. Find the coordinates of all intersectionsof the circle with an origin-centered square of side length 4 whose sides are parallel to the coordinateaxes.

Solution: (±1.5,±2), (±2,±1.5)The square consists of all points (x, y). with −2 ≤ x, y ≤ 2 and at least one of x and y equal to 2 or-2. The points on the circle are all points (x, y) with x2 + y2 = 6.25. So, the intersection of these twois (±2,±1.5) and (±1.5,±2).

25. Write 2010(7017)!(1708)!

as a decimal in base 6. The subscript indicates the base in which the number is

written (i.e., 156 is 15 base 6, so 156 = 1110)

Solution: 0.16 or 0.1

The easiest way to solve this problem is to convert everything into base 10.

7017 = 7× 171 + 0× 170 = 119

1708 = 1× 82 + 7× 81 + 0× 80 = 120

2010(7017)!

(1708)!= 20

119!

120!=

1

6

Converting to base 6, 16 = 0.16.

Page 17: Smt Solutions

26. Jacques is building an igloo for his dog. The igloo’s inside and outside are both perfectly hemispherical.The interior height at the center is 2 feet. The igloo has no door yet and contains 254

2187π cubic yardsof hand-packed snow. What is the circumference of the igloo at its base in feet?

Solution: 143π

Let R be the radius of the igloo at its base, in feet. Then, the amount of snow in the igloo is, in cubicfeet, 12

(43πR

3 − 43π · 23)= 23π(R

3 − 8). This is equal to 2542187π cubic yards, or 25481 = 33 · 2542187 cubic

feet. Therefore, 23π(R3 − 8) = 254

81 π, so R3 = 34327 , and R = 7

3 . So, the circumference of the igloo is2πR = 14

3 π feet.

27. How many solutions are there to x7 + y7 = z7 with x, y, z real?

Solution: ∞For any real x, y, x7 + y7 is real, so x7 + y7 has a 7-th root. If we let z be this root, then (x, y, z) is asolution to x7 + y7 = z7. So, for any x, y, there is a solution. There are infinitely many x, y, so theremust be infinitely many solutions.

28. Find all prime factors of 318 − 218.

Solution: 5, 7, 19, 577, 1009

318 − 218 = (39 − 29)(39 + 29) = (33 − 23)(36 + 33 · 23 + 26)(33 + 23)(36 − 33 · 23 + 26)

= 19 · 1009 · 35 · 577 = 5 · 7 · 19 · 577 · 1009

29. If a = 1 and a√10000a = 10a, find a. The subscript indicates the base in which the number is written.

Solution: 4

10000a = a4 and 10a = a, soa√a4 = a. Thus, a4/a = a, and 4/a = 1, so a = 4.

30. Suppose 100 students have at least one of math, applied math, or statistics as one of their majors.There are 24 statistics majors, 46 applied math majors, and 55 pure math majors. There are 14 whoare at least pure & applied math, 10 at least applied math & statistics and 7 at least statistics & puremath. How many triple majors (people with all three majors) are there, if any?

Solution: 6

Using the principle of inclusion-exclusion, the number of triple majors is given by 100 − (46 + 24 +55) + (14 + 10 + 7) = 6.

31. The sum of 3 real numbers is 0. If the sum of their cubes is πe, what is their product?

Solution: πe

3

Let x, y, and z be the 3 real numbers. Then x+ y + z = 0, so z = −(x+ y). Also,

πe = x3 + y3 + z3 = x3 + y3 − (x+ y)3

= −3x2y − 3xy2 = −3xy(x+ y) = 3xyz

So, xyz = 13πe.

32. Given a real-valued function f(x) =√

x+3|2x+6| , what is the least integer value which lies in the domain

of the function?

Solution: −2We have x ≥ −3, as the value under the radical must be nonnegative, and the absolute value is nonnegative, so x + 3 ≥ 0. But, the denominator must be non-zero, so we must exclude x = −3, as then2x+ 6 = 0. Hence, the least integer value is -2.

Page 18: Smt Solutions

33. E is a point in the interior of rectangle ABCD. AB = 6, triangle ABE has area 6, and triangle CDEhas area 12. Find (EA)2 − (EB)2 + (EC)2 − (ED)2.

Solution: 0Let l be the line through E perpendicular to AB and CD. Let F and G be the intersections of l withAB and CD, respectively. Then, AF = DG, call this a, and BF = CG, call this b. Since triangleABE has area 6, EF = 2. Since triangle CDE has area 12, EG = 4.

A B

CD

E

F

G

a b

a b

2

4

By the Pythagorean Theorem, we have the following equalities:

a2 + 4 = (EA)2, b2 + 4 = (EB)2, b2 + 16 = (EC)2, a2 + 16 = (ED)2

Thus, (EA)2 − (EB)2 + (EC)2 − (ED)2 = 0.

34. What is the radius of the incircle of a right triangle with legs of lengths 7 and 24?

Solution: 3

Let ABC be a right triangle with AB = 24 and BC = 7. Then, AC = 25. Let x be the radius of theincircle of ABC. Let D, E, and F be the points on AB, BC, and AC, respectively, that are tangentto the incircle. Then, AD = AF = 24− x and CE = CF = 7− x. So, AC = 31− 2x = 25, and x = 3.

35. If logAB + logB A = 3 and A < B, find logB A.

Solution: 3−√5

2

Let x = logB A. Then Bx = A, so B = A1x . Thus, logAB = 1

x. So, x + 1

x= 3. Multiplying through

by x, x2 + 1 = 3x, so x2 − 3x+ 1 = 0. By the quadratic formula,

x =3±√5

2

Since A < B, logB A < 1, so

x =3−√5

2.

36. ABC is an equilateral triangle with edge of length 256. Let n be the maximum number of non-overlapping equilateral triangles of edge length 1

2 that can be fit into ABC. What is the primefactorization of n?

Solution: 218

For a triangle of side length n, we can fit in 4 triangles with sides of length n2 (maybe draw a figure).

25629 = 1

2 , so we can do this 9 times. Each time, the number of triangle increases 4-fold. 49 = 218.

37. If x2 + 1x2

= 7, find∣∣x3 + 1

x3

∣∣.Solution: 18

Observe that (x + 1x )2 = x2 + 2 + 1

x2 = 9. so∣∣x+ 1

x

∣∣ = 3. Cubing,∣∣x3 + 3x+ 3

x +1x3

∣∣ = 27. Now,x3+ 1

x3has the same sign as 3x+ 3

xbecause if x is positive, both are positive, and if x is negative, both

are negative. So,∣∣x3 + 3x+ 3

x+ 1x3

∣∣ = ∣∣x3 + 1x3

∣∣+ 3∣∣x+ 1

x

∣∣ = 27. Since∣∣x+ 1

x

∣∣ = 3,∣∣x3 + 1

x3

∣∣ = 18.

Page 19: Smt Solutions

38. Eleven pirates find a treasure chest. When they split up the coins in it evenly among all the pirates,they find that there are 5 coins left. They throw one pirate overboard and split the coins again, onlyto find that there are 3 coins left over. So, they throw another pirate over and try again. This time,the coins split perfectly. What is the least number of coins there could have been?

Solution: 423

Let n be the number of coins in the chest. Then, n ≡ 5 (mod 11) (this means that the remainder ofn when divided by 11 is 5), n ≡ 3 (mod 10), and n ≡ 0 (mod 9). The Chinese Remainder Theoremgives an elegant solution to this problem, or you can solve it as follows. Since n ≡ 3 (mod 10),we know that the last digit of n must be 3. Also, since n ≡ 0 (mod 9), n is divisible by 9, son = 9q for some q. Since n ends in a 3, q must end in a 7. That means that q looks like 10p+ 7, son = 9(10p+ 7) = 90p+ 63. Now, you can just try p = 1, 2, . . ., and you’ll find that p = 4 works.

Page 20: Smt Solutions

2001 Stanford Math Tournament

Team Test

1. ABCD is a square with sides of unit length. Points E and F are taken on sides AB and AD respectivelyso that AE = AF and the quadrilateral CDFE has maximum area. What is this maximum area?

Solution: 5/8

Let x = AE. Then, BE = 1 − x, so the area of CDFE is 1 − x2

2 − 1−x2 = 1 − x2−x+1

2 , which ismaximized when x2 − x + 1 is minimized. x2 − x + 1 is minimized at x = 1

2 . Plugging this into ourformula for the area of CDFE, we find that the maximum area of CDFE is 58 .

2. How many positive integers between 1 and 400 (inclusive) have exactly 15 positive integer factors?

Solution: 3

In general, if a number has prime factorization pn11 pn22 · · · pnkk , then its positive integer factors are

exactly those of the form pj11 pj22 . . . pjkk where 0 ≤ ji ≤ ni for each i between 1 and k, inclusive. In

particular, there are ni + 1 choices for ji, so there are (n1 + 1)(n2 + 1) · · · (nk + 1) positive integerfactors. In our case, since 15 = 1 · 15 = 3 · 5 are the only factorizations, either n1 = 0 and n2 = 14 orn1 = 2 and n2 = 4. So, a positive integer has exactly 15 positive integers if and only if it is of the formp14 for some prime p or of the form p21p

42 for two distinct primes p1 and p2. 214 > 400, so there are

no numbers of the first form between 1 and 400. The only ones of the second form that are between1 and 400 are 32 · 24 = 144, 52 · 24 = 400, and 22 · 34 = 324. Therefore, there are 3 positive integersbetween 1 and 400, inclusive, that have exactly 15 positive integer factors.

3. Find the 2000th positive integer that is not the difference between any two integer squares.

Solution: 7998

First, we will try to find all numbers that can be expressed as a difference between two integer squares.For any integer n, (n+1)2−n2 = 2n+1, so any odd number can be expressed as the difference betweentwo integer squares. Similarly, (n+2)2 − n2 = 4n+ 4 = 4(n+ 1), so any number divisible by 4 can beexpressed as the difference between two integer squares.

What about the other numbers, the ones of the form 4n+ 2? It is easy to see that none of these canbe the difference between two integer squares. Assume that it could; then x2 − y2 = 4n+ 2 for somex and y. Then, either x and y are both even, or they are both odd, since their difference is even.However, then x2 − y2 would be divisible by 4, while 4n + 2 is not. So, the integers that cannot beexpressed as the difference between two integer squares are exactly those of the form 4n+2. The first

is 4(0) + 2 = 2, so the 2000th one is 4(1999) + 2 = 7998.

4. For what values of a does the system of equations

x2 = y2, (x− a)2 + y2 = 1

have exactly 2 solutions?

Solution: ±√2We can plot this system to get a rough idea of what it looks like. x2 = y2 is the same as x = ±y, ora pair of perpendicular lines. (x− a)2 + y2 = 1 is a circle with center (a, 0) and radius 1. We can seethat the system will have exactly 2 solutions if and only if the circle is tangent to both lines:

O A

Page 21: Smt Solutions

Let’s label the origin O, the center of the circle A = (a, 0), and the two points of tangency B andC. Then, OBAC is a rectangle, and AB = AC = 1. Thus, OBAC must be a square, and AO, itsdiagonal, must have length

√2. Thus, a =

√2. Now, of course, the picture is exactly the same if the

circle is on the other side of the origin, so a = −√2 is also a possible solution. Thus, the values of agiving exactly 2 solutions are ±√2.

5. What quadratic polynomial whose coefficient of x2 is 1 has roots which are the complex conjugates ofthe solutions of x2 − 6x+ 11 = 2xi− 10i? (Note that the complex conjugate of a+ bi is a− bi, wherea and b are real numbers.)

Solution: x2 + (−6 + 2i)x+ (11− 10i)Let a, b be the solutions of x2−6x+11 = 2xi−10i. Then, a and b are the roots of x2−(6+2i)x+(11+10i),so x2 − (6 + 2i)x + (11 + 10i) = (x − a)(x − b) = x2 − (a + b)x + ab. Therefore, a + b = 6 + 2i andab = 11 + 10i.

Let a, b the complex conjugates of a, b, respectively. Then, the quadratic polynomial with one as thecoefficient of x2 and a, b as roots is (x − a)(x − b) = x2 − (a + b)x + ab = x2 − (a+ b)x + ab =x2 − (6 − 2i)x+ (11− 10i) = x2 + (−6 + 2i)x+ (11− 10i) = 0.

6. Find the least n such that any subset of {1, 2, . . . , 100} with n elements has 2 elements with a differenceof 9.

Solution: 55

Let A = {18a+ 1, 18a+ 2, . . . , 18a+ 9 : a = 0, 1, 2, 3, 4, 5}. Then, notice that A has 54 elements, andno 2 elements have a difference of 9. So, we know n ≥ 55. Now, we just need to convince ourselvesthat any subset with 55 elements must have 2 elements with a difference of 9. To do this, take thesets {x, x + 9} for x = 1, 2, . . . , 91. Notice that there are 91 such sets. The numbers 1, 2, . . . , 9 and92, . . . , 100 each fall into exactly one of the set, while all other numbers fall into two of them. So, thereare 18 numbers that fall into exactly one set, and then there are 73 other sets. That means that, in aset of 55 elements, at least 37 fall into two sets. As 37 · 2 + 18 = 92 > 91 (the number of sets), by thepigeonhole principle, two of them fall in the same set, and thus have a difference of 9.

7. The median to a 10 cm side of a triangle has length 9 cm and is perpendicular to a second median ofthe triangle. Find the exact value in centimeters of the length of the third median.

Solution: 3√13

Call the triangle ABC, where AB = 10, BC = 2x, and CA = 2y. Let D, E, and F be the midpoints ofAB, BC, and CA, respectively. Let G be the centroid (the point where the three medians intersect).

A

B

C

D

EF

G

55

x

xy

y

We are given that CD = 9. One property of the centroid is that CGGD = AGGE = BG

GF = 2. So, CG = 6and GD = 3. We are given that CD is perpendicular to another median, say AE. Let a = GE; thenAG = 2a. By the Pythagorean Theorem, a2 + 36 = x2, (2a)2 + 36 = (2y)2, and 9 + (2a)2 = 25. Thelast tells us that a = 2, so then x = 2

√10 and y =

√13.

Let θ = ∠BCA. By the law of cosines with triangle ABC, 100 = (2y)2 + (2x)2 − 2(2x)(2y) cos θ,so xy cos θ = 100−52−160

−8 = 14. By the law of cosines with the triangle BCF , BF 2 = y2 + (2x2) −2y(2x) cos θ = 13 + 160− 4 · 14 = 117. So, BF = 3

√13.

Page 22: Smt Solutions

8. Janet and Donald agree to meet for lunch between 11:30 and 12:30. They each arrive at a randomtime in that interval. If Janet has to wait more than 15 minutes for Donald, she gets bored and leaves.Donald is busier so will only wait 5 minutes for Janet. What is the probability that the two will eattogether? Express your answer as a fraction.

Solution: 43144

We will solve this problem using a graph. Let the x-axis be time for Janet, and let the y-axis be timefor Donald. Then, the region we are interested in is the square with both x and y between 11:30 and12:30. Since Janet will wait no more than 15 minutes for Donald, we have x ≤ y + 15. Since Donaldwill wait no more than 5 minutes for Janet, y ≤ x+ 5.

12:3011:30

12:30

Don

ald

Janet

The area of the whole square is 602, the area of the upper triangle is 452

2 , and the area of the lower

triangle is 552

2 , so the probability that the two will eat together is602− 4522 − 55

2

2

602 = 43144 .

9. What is the minimum number of straight cuts needed to cut a cake in 100 pieces? The pieces do notneed to be the same size or shape but cannot be rearranged between cuts. You may assume that thecake is a large cube and may be cut from any direction.

Solution: 9

First, we will find the number of regions that n lines split a plane into. Notice that 0 lines split theplane into 1 region, 1 line splits the plane into 2 regions, 2 lines splits it into 4 regions, etc. In general,the k-th line can intersect each of the first k − 1 lines exactly once, creating k more regions. So, thenumber of regions that n lines split a plane into is

1 +n∑k=1

k =n(n+ 1)

2+ 1

Now, cutting the cake is the same idea, only in 3 dimensions instead of a plane. Again, 0 planes splitthe cake into 1 piece, 1 plane gives 2 pieces, 2 planes give 4 pieces, etc. Notice that the k-th plane canintersect each of the first k − 1 planes exactly once, but now the intersection is a line. Then, by the

planar case, these lines divide the new plane into (k−1)k2 +1 regions; each of these planar regions splits

one piece of the cake into two, so we’ve added (k−1)k2 +1 pieces. So, the number of regions that n linessplit a cube into is

1 +

n∑k=1

((k − 1)k

2+ 1

)= 1 +

1

2

n∑k=1

k2 − 1

2

n∑k=1

k +

n∑k=1

1

= 1 +1

2

n(n+ 1)(2n+ 1)

6− 1

2

n(n+ 1)

2+ n

= n+ 1 +2n3 + 3n2 + n

12− n2 + n

4

=2n3 + 10n+ 12

12=

n3 + 5n+ 6

6

Then, the least n so that n3+5n+66 ≥ 100 is 9.

10. You know that the binary function � takes in two non-negative integers and has the following properties:

Page 23: Smt Solutions

• 0 � a = 1

• a � a = 0

• If a < b, then a � b = (b− a)[(a− 1) � (b− 1)].

Find a general formula for x � y, assuming that y ≥ x > 0.

Solution: (y − x)xWe just need to check that this satisfies the three given properties. First, 0 � a = (a− 0)0 = 1. Second,a�a = (a−a)a = 0. Finally, if a < b, then a�b = (b−a)a = (b−a)[(b−a)a−1] = (b−a)[(a−1)�(b−1)].

11. Christopher and Robin are playing a game in which they take turns tossing a circular token of diameter1 inch onto an infinite checkerboard whose squares have sides of 2 inches. If the token lands entirelyin a square, the player who tossed the token gets 1 point; otherwise, the other player gets 1 point. Aplayer wins as soon as he gets two more points than the other player. If Christopher tosses first, whatis the probability that he will win? Express your answer as a fraction.

Solution: 12

First, notice that the player tossing wins a point if the center of the coin lands in the center 1× 1 bysquare of each 2× 2 square of the checkerboard. So, the player tossing wins a point with probability14 . Now, a player wins when the other player has n points and he has n+ 2, making a total of 2n+ 2tosses. We can think about the tosses in pairs; the first two are a pair, the second two are a pair,etc. The first player to win one of these pairs also wins the game, so the probability that Christopherwins the game in 2n+2 turns is the probability that, in the first n pairs, Christopher gets 1 point andRobin gets 1 point, and in the last pair, Christopher wins both points.

Christopher tosses first in each pair, so his probability of winning the first toss is 14 , and his probabilityof winning the second toss by Robin not winning is 34 . So, the probability that he wins the first tossand loses the second toss is 14 · 14 = 1

16 . The probability that he loses the first toss and wins the secondis 34 · 34 = 9

16 . So, the probability that Christopher and Robin each get 1 point in the pair is 1016 . The

probability that this happens for the first n pairs is(1016

)n. The probability that Christopher wins the

last pair is 14 · 34 . So, the probability that Christopher wins in 2n + 2 turns is(1016

)n 316 . This means

that, overall, the probability that Christopher wins is

∞∑n=0

(10

16

)n3

16=

3

16

[1 +

10

16+

(10

16

)2+

(10

16

)3+ . . .

]

This is a geometric series with sum3

16· 1

1− 1016=

1

2

(Notice that the fact that the person tossing wins with probability 14 is not important – no matter

what that probability is, the overall probability of winning in such a game is still 12 !!)

12. A binary string is a string consisting of only 0’s and 1’s (for instance, 001010, 101, etc.). What isthe probability that a randomly chosen binary string of length 10 has 2 consecutive 0’s? Express youranswer as a fraction.

Solution: 5564

We will first find the number of binary strings of length 10 that don’t have 2 consecutive zeros. LetCn bethe set of binary strings of length n that don’t have 2 consecutive zeros. Let An be the set of all stringsin Cn that end in 1, and let Bn be the set of all strings in Cn that end in 0. Notice that Cn = An∪Bn.Now, if a string is in An+1, then, the first n letters must be a string in Cn; conversely, appending a 1 toa string in Cn gives a string in An+1. So, |An+1| = |Cn| = |An|+ |Bn|. Similarly, a string is in Bn+1 iffthe first n letters are in An, so |Bn+1| = |An|. Thus, |An+2| = |An+1|+ |Bn+1| = |An+1|+ |An|, whichgives us a Fibonacci sequence which starts with |A1| = 1, |A2| = 2. It’s easy to find that |A10| = 89 and|B10| = 55, so |C10| = 89+55 = 144. Therefore, there are 210−144 = 1024−144 = 880 strings without2 consecutive 0’s. So, the probability that a randomly chosen string of length 10 has 2 consecutive 0’sis 8801024 =

5564 .

Page 24: Smt Solutions

13. You have 2 six-sided dice. One is a normal fair die, while the other has 2 ones, 2 threes, and 2 fives.You pick a die and roll it. Because of some secret magnetic attraction of the unfair die, you have a75% chance of picking the unfair die and a 25% chance of picking the fair die. If you roll a three, whatis the probability that you chose the fair die?

Solution: 17

The probability of picking the fair die is 14 , and the probability of rolling a 3 with it is 16 . So, theprobability of picking the fair die and rolling a 3 with it is 1

24 . The probability of picking the unfairdie and rolling a 3 with it is 34 · 13 = 1

4 . So, the probability that we chose the fair die is

124

124 +

14

=1

7

14. Find the prime factorization of∑1≤i<j≤100 ij.

Solution: 3 · 52 · 11 · 101 · 151We will find

∑1≤i<j≤n ij for any n. Since (1 + 2 + . . . + n) = n(n+1)

2 , we square and get that

(1 + 2 + . . . + n)2 =(n(n+1)2

)2. Since 2

∑1≤i<j≤n ij = (1 + 2 + . . .+ n)2 − (12 + 22 + . . . + n2), we

have ∑1≤i<j≤n

ij =1

2

[n2(n+ 1)2

4− n(n+ 1)(2n+ 1)

6

]

=3n4 + 6n3 + 3n2 − 2(2n3 + 3n2 + n)

24

=3n4 + 2n3 − 3n2 − 2n

24=

n(3n3 + 2n2 − 3n− 2)

24

=n(3n+ 2)(n2 − 1)

24=

(n− 1)n(n+ 1)(3n+ 2)

24

So, for n = 100, we find that the sum is 99·100·101·30224 = 33 · 25 · 101 · 151 = 3 · 52 · 11 · 101 · 151.15. Let ABC be an isosceles triangle with ∠ABC = ∠ACB = 80◦. Let D be a point on AB such that∠DCB = 60◦ and E be a point on AC such that ∠ABE = 30◦. Find ∠CDE in degrees.

Solution: 30◦

Let θ = ∠CDE.A

D

E

5060

70

110

20

2030

40

50

C B

110

θ

Since we’re only worried about angles in this problem, we can just pretend that AD = 1. Then, sinceADC is an isosceles triangle with side 1 and base angle 20◦, AC = 2 cos 20◦. Since ABC is an isoscelestriangle with vertex angle 20◦ and side 2 cos 20◦, BC = 4 cos 20◦ sin 10◦. Notice that ECB is isosceles,so EC = BC = 4 cos 20◦ sin 10◦. Also, ADC is isosceles, so DC = AD = 1.

Now, let’s look at the triangle EDC. Notice that ∠DEC = 110− θ. By the law of sines,

sin θ

EC=

sin(160◦ − θ)

DC

Page 25: Smt Solutions

so

sin θ

4 cos 20◦ sin 10◦= sin(160◦ − θ)

= sin(20◦ + θ)

= sin 20◦ cos θ + cos 20◦ sin θ

Then,

cos θ sin 20◦ = sin θ

(1

4 cos 20◦ sin 10◦− cos 20◦

)= sin θ

(1− 4 cos2 20◦ sin 10◦

4 cos 20◦ sin 10◦

)So,

1

tan θ=

1− 4 cos2 20◦ sin 10◦

4 cos 20◦ sin 10◦ sin 20◦

=1− 4 cos2 20◦ sin 10◦

2 sin 40◦ sin 10◦since sin(2φ) = 2 sinφ cosφ

=cos 10◦ − 4 cos2 20◦ sin 10◦ cos 10◦

2 sin 40◦ sin 10◦ cos 10◦

=cos 10◦ − 2 cos2 20◦ sin 20◦

sin 40◦ sin 20◦

=sin 80◦ − cos 20◦ sin 40◦

sin 40◦ sin 20◦=

2 sin 40◦ cos 40◦ − cos 20◦ sin 40◦

sin 40◦ sin 20◦

=2 cos 40◦ − cos 20◦

sin 20◦=

2 cos(30◦ + 10◦)− cos(30◦ − 10◦)sin(30◦ − 10◦)

=2(cos 30◦ cos 10◦ − sin 30◦ sin 10◦)− (cos 30◦ cos 10◦ + sin 30◦ sin 10◦)

sin 30◦ cos 10◦ − cos 30◦ sin 10◦

=

√32 cos 10◦ − 32 sin 10◦12 cos 10

◦ −√32 sin 10◦

=√3

So, tan θ =√33 , which means that θ = 30◦.